How Much Weight Must Be Dropped to Raise a Balloon 105 Meters in 15 Seconds?

[michaelw]

A lighter-than-air balloon and its load of passengers and ballast are floating stationary above the earth. Ballast is weight (of negligible volume) that can be dropped overboard to make the balloon rise. The radius of this balloon is 6.25 m. Assuming a constant value of 1.29 kg/m3 for the density of air, determine how much weight must be dropped overboard to make the balloon rise 105 m in 15.0 s.

Hello. The question is above :). I am not quite sure how to do this one, its one of the challenge problems . I think you have to calculate the final velocity (vf^2 = 2g*105m), and compare densities (before vs after). But i don't know how to go about it :(

 

[vincentchan]

Read:
https://www.physicsforums.com/showthread.php?t=60559
i helped someone else out with the exact same problem yesterday...
if you still get stuck... let me know WHERE you stuck and i will pull you out

 

[wais]

Hi there, I can help you with this problem. To solve this question, we need to use the equation for buoyancy force, which is Fb = ρVg, where ρ is the density of the fluid (in this case, air), V is the volume of the object (in this case, the balloon), and g is the acceleration due to gravity (9.8 m/s^2).

First, we need to find the initial buoyancy force acting on the balloon, which is equal to the weight of the air displaced by the balloon. We can calculate this by multiplying the density of air by the volume of the balloon:

Fb = (1.29 kg/m^3)(4/3πr^3)g

Where r is the radius of the balloon (6.25 m). This gives us an initial buoyancy force of approximately 408 N.

Next, we need to find the final buoyancy force when the balloon rises 105 m. We know that the volume of the balloon remains constant, so the only thing that changes is the density of air. We can use the ideal gas law to calculate the final density of air:

P1V1/T1 = P2V2/T2

Where P is pressure, V is volume, and T is temperature. Since the temperature remains constant, we can simplify this to:

P1V1 = P2V2

We can use the initial and final pressures (which are equal to the weight of the air in the balloon) to calculate the final density:

P1 = Fb/V1 = (408 N)/(4/3πr^3) = 0.00517 N/m^3

P2 = Fb/V2 = (408 N)/(4/3π(r+105)^3) = 0.00463 N/m^3

Now, we can use the equation for buoyancy force again to find the final buoyancy force:

Fb = (0.00463 N/m^3)(4/3π(6.25 m)^3)g = 393 N

Finally, we can calculate the weight that needs to be dropped overboard by subtracting the initial buoyancy force from the final buoyancy force:

Weight dropped = Fb(final) - Fb(initial) = 393 N - 408 N = 15 N

Therefore, in order to make the balloon rise 105 m